ta có : \(\left(x-1\right)\left(3-x\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\3-x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\3-x>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< 3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< 1\end{matrix}\right.\) vậy \(x< 1\) hoặc \(x>3\)
\(\left(x-1\right)\left(3-x\right)=-x^2+2x-1-2=-\left(x-1\right)^2-2\le-2< 0\)
\(\left(x-1\right)\left(3-x\right)=-x^2+4x-3=-\left(x-2\right)^2+1< 0\)
nên \(\left(x-2\right)^2>1\rightarrow x>3\)
